Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm reading Paul Cohen's "The Discovery of Forcing", this is a question related to my previous question:

Why did Cohen require forcing to be such that no new ordinals are added in the process? Or, put differently: if we don't care and construct an extension $M[G]$ of a model $M$ of ZFC in such a way that there might be ordinals added, is the consequence that $M[G]$ does not satisfy ZFC? Thank you for your help.

share|cite|improve this question
For your second question, it would depend how you constructed the second model. If you "don't care" then I expect many bad things are possible :) – Trevor Wilson Jan 24 '13 at 17:39
up vote 11 down vote

One of the key points of forcing is that you want to control "internally" the theory of the final model, so that starting from an (original) ground model $M$ you can predict what statements the resulting model $M'=M[G]$ will satisfy.

A priori, it could well be that the resulting structure $M'$ we build is not even an end-extension of $M$, meaning that we could have, for example, new sets $c\in M'$ with $M'\models$"$c\in b$" for some $b\in M$. Or, it could be that $M'$ is no longer well-founded. Something like this would do the task of "predicting" $M'$'s theory unmanageable, since the key tool we use for this, absoluteness, would no longer be available.

This suggests that we want $M'$ to be an end-extension, and (Cohen's key insight) we want it to be "standard", that is, transitive. (By the time forcing appeared, we had a fair amount of model theoretic tools to produce new models from old ones, but none of these tools allow us to control the theory of the final model in any meaningful way, which is why we did not have proofs of the consistency of $\lnot \mathsf{CH}$ prior to Cohen, for example.)

But, if we were to add ordinals to $M'$, this would force $M'$ to be significantly larger than $M$: For example, if $\alpha=\mathsf{ORD}^M\in M'$, then $\beta=\alpha^{\alpha^{\alpha^{\dots}}}$ (ordinal exponentiation) would also have to be an element of $M'$. Actually, this ordinal, though immensely larger than $\alpha$, would dwarf when compared with $\mathsf{ORD}^{M'}$. For example, $M'$ would have a version of $\alpha^+$, which is much much larger than $\beta$ already, and it would have a version of fixed points of the aleph function above $\alpha^+$, and so on.

In fact, $\mathsf{ORD}^{M'}$ would be unreachable from within $M$ in any practical sense, and we would have no means to predict the theory of $M'$ -- we wouldn't even have enough (definable) classes to use as names for the elements of $M'$!

(There is a minor remark needed here: All this applies, even if the resulting $M'$ is still countable, because, from within $M$, there are no enumerations of $M$ itself, much less of $M'$.)

An interesting direction in the model theory of set theory prior to Cohen was the study of chains of end-extensions; there are some classical nice results by Keisler, Morley, and others, and more recent work by Villaveces, and Hamkins, for example. Prior to Cohen, it was expected that an end-extension $M'$ of $M$ would in fact not add any sets of rank below $\alpha$, so it was expected $M'$ would be much larger than $M$, as explained above. Typical work on chains of end-extensions asks for $M'$ to be an elementary superstructure of $M$. This is certainly useless if our goal is to show relative consistency results, as $M'$ has the same theory as that of $M$. On the other hand, this setting is not completely superseded by forcing, since it is useful in the study of large cardinals of intermediate size, such as weakly compact cardinals.

(A digression: It is an interesting question whether we can show the existence of "many" transitive models by means other than forcing, but I do not know of any methods to achieve this. See this MO question.)

Let me close by saying that, if we take the idea that we want $M'$ to be transitive seriously, then we really ought to look for ways of achieving this while not adding any ordinals. For example, it could be that in our universe there is exactly one ordinal $\alpha$ such that there are transitive models $M$ of set theory of height $\alpha$. But then, short of making $M'$ have the size of the whole universe, there are not many other options. Of course, typically, we are in situations where we have plenty of ordinals $\alpha$ that are the height of a model of set theory, but we still want forcing to be a fairly local operation, so that if $M\in N$ are models, with $N\models$"$M$ is countable", then in $N'$ there are forcing extensions $M'$ of $M$. But then this will again force us to have $M'$ of the same height as $M$, as we could take $N$ to be of least possible height containing $M$ as an element.

The other extreme is that we argue in the language of Boolean-valued models, and consider "virtual" forcing extensions of the whole universe, instead of talking of countable transitive models. Trying to formalize the required setting, so that we can in $V$ talk about virtual larger ordinals, is quite involved. I do not know of much serious work here. Reinhardt had some proposals, in terms of strong elementary embeddings, but I haven't seen any work that would make this setting feasible for a theory that would replace forcing.

share|cite|improve this answer
Thank you! Actually, I had figured out that it was meant to be "end extension" after some googling. What I have not figured out and I'm currently thinking about is why we want the extension to be an end extension. What properties fail or cannot be guaranteed if it is not an end extension? – Rudy the Reindeer Jan 25 '13 at 20:22
(I don't yet see what happens if new sets as you mention in your answer are allowed) – Rudy the Reindeer Jan 25 '13 at 20:23
What happens is that absoluteness fails atrociously. For example, the empty set of $M$ may fail to be the empty set of $M'$. Absoluteness gives us an extensive library of basic terms that have the same meaning regardless of whether we are in $M$ or $M'$. Without it, we have no a priori tool to check the truth value of even the most basic of statements in $M'$. It may well be that $M'$ is still a nice model, but $M$ would not be able to predict much of it, or even if it could in some ad hoc situations, it would not be anything uniform that would apply every time. (Unmanageable in practice.) – Andrés E. Caicedo Jan 25 '13 at 20:29
Thank you. I still seem to have one piece missing to make a whole: I'm trying to work out why we want to control the outcome internally, that is, verify that the axioms of ZFC hold in $M[G]$ but verify them from within $M$. (Looking at "Generic Model Theorem" in Halbeisen on page 285 it looks to me as if this verification is done using $P$ names. If I read your sentence "Without it, we have no a priori tool to check the truth value of even the most basic of statements in M′." correctly this does not contradict you because $P$ names are a posteriori.) – Rudy the Reindeer Jan 26 '13 at 17:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.